Question

Find the limit:

Answer 1

\[\LARGE \lim_{x \rightarrow 0} \frac{\sin^2x}{sinx^2}\]

Answer 2

multiply both top and bottom by x^2

Answer 3

We can use L'hopital's rule here can't we? (Finally officially "learned" it)

Answer 4

Yes, you could. Either way works

Answer 5

without using L;hospital rule, we have:
(sinx / x) (sinx / x) (x^2 / sin(x^2))


each part has a limit of 1. so 1*1*1 = 1

Answer 6

but if you wanna do L'hospital then just take the derivative of top and bottom

Answer 7

Thank you :)